Early Atomic Models. Atoms: the smallest particle of an element that retains the properties of that element.

 Atoms: the smallest particle of an element

that retains the properties of that element.

 (Greek: atomos = indivisible)

 Democritus (Greek teacher in the 4th

century BC)

 First suggested the idea that atoms

existed

Law of Conservation of Mass

 Definition: mass cannot be created or destroyed

John Dalton

 English school teacher

1. All elements are composed of tiny

indivisible particles called atoms.

2. Atoms of the same element are

identical. The atoms of any one

element are different from those of

any other element.

3. Atoms of different elements can combine with one another in simple whole number ratios to form compounds.

 H₂O C₁₂H₂₂O₁₁ NOT H_2.5O_¾

Most of Dalton’s Atomic Theory is accepted

 One major revision includes that idea that

atoms are indivisible….

There are 3 parts to an atom….

1. electrons 2. protons 3. neutrons

Negatively charged subatomic particles

 J.J. Thomson discovered in 1897

 Passed a electric current through gases at low pressures

called a “Cathode Ray Tube”

 Noticed the surface of the tube directly opposite the

cathode glowed.

 Why? Opposites attract and the electrons were attracted

to the positive ends and lights up!

 Cathode Ray Tube

Cathode rays are identical regardless of the element

 Therefore all elements must have electrons!

Other important findings:

 Atoms are electrically neutral, so they must

contain a positive charge to cancel it out

 Since electrons are so small, atoms must contain

J.J. Thomson – plum-pudding model

 e- are spread evenly though out the positive

charge of the rest of the atom

 Ms. Agostine’s “mint chocolate chip ice

Robert Millikan (1868-1953)



Found quantity of charge in 1

electron (e

₎



Also determined the ratio of the

charge to the mass of 1 e

-

Calculated the mass of 1 e

Ernest Rutherford (1911)

 nucleus of the atom is positively

charged

 Most  particles go straight through

 Positively charged  particles deflect off of the

positively charged nucleus(~1/8,000)

 “…it was as if you fired a 15-inch [artillery]

shell at a piece of tissue paper and it came back and hit you.”

 Nucleus was very small

 If a nucleus were a marble

 Neutrons (no)

 Subatomic particles

with no charge

 Discovered by Sir

James Chadwick

 Mass is nearly the

same as a proton

Atomic Number : the number of protons in the

nucleus of an atom of an element

Atoms are electrically neutral

 Tells how many electrons there are also! Periodic Table

#1 – Hydrogen: has 1 p+ _{and 1 e}-

Mass Number – total number of protons and

neutrons in a nucleus

# of neutrons = mass # - atomic # = (# p+ _{+ # n}o_{) - (# p}+₎

Ex) Beryllium – 9

 Hyphen notation: The number “9” is the

mass number

Definition – atoms that have the same number of protons but different numbers of neutrons

 Different types of the same element  Ex) Carbon – has 3 isotopes

 Carbon – 12  Carbon – 13  Carbon – 14

All have the same # of p+

 If not, it would be a different element  All have 6 protons

Hydrogen-1: 1 p+ _{and 0 n}o

 Relative abundance = 99.985 %

 Commonly called normal “hydrogen”

Hydrogen-2: 1 p+ _{and 1 n}o

 Relative abundance = 0.015%

 Commonly called heavy hydrogen or “deuterium”

Hydrogen-3: 1 p+ _{and 2 n}o

Definition – weighted average mass of the atoms in a naturally occurring sample of the element

Carbon-12 = 98.89 % abundant Carbon-13 = 1.11% abundant

Formula:

Atomic = relative • mass # + relative • mass # + mass abund. abund.

Repeats for as many isotopes as exist for that element….

Units: atomic mass unit (amu): defined as exactly 1/12 the mass of a carbon-12 atom

 1 amu = approximately the mass of 1 proton

 amu’s are used so you don’t have to use scientific notation

Sample Problem:



Chlorine has 2 isotopes:

chlorine-35 which is 75.77%

abundant and chlorine-37

35 _{Cl = 75.77% abundant = 0.7577 rel. abund.} 37 _{Cl = 24.33% abundant = 0.2433 rel. abund.}

Atomic mass =

= (35 amu x 0.7577) + (37 amu x 0.2433) = (26.5195 amu) + (9.0021 amu)

= 35.5 amu